Two pith balls of equal mass M and equal charge q are suspended from the same point on long massless threads of length L as shown in the figure above. If k is the Coulomb's law constant, then for small values of , the distance d between the charged pith balls at equilibrium is

Electromagnetism}Forces

Sum of the forces for one of the mass in the x (horizontal) and y (vertical) directions gives,

For small angles, . From the geometry, one can deduce that .

Thus, the x equation yields (since from the y equation for small angles). This is choice (A).

Alternate Solutions

cjohnson4152013-08-19 01:40:11

Here's an alternative solution that skips the trig altogether:

Due to small , we can consider each as a simple pendulum, a special case of the simple harmonic oscillator:

Thus , which is the restoring force constant (), so:

The rest is the same algebra, just a different/potentially quicker/potentially prettier way to arrive at the force balance.

Dimensional analysis helps!
Since is in N, will have units of
Mg has units of N
so will have units of
This leaves only A and B. Knowing that solving for will give half of d, take the solution with an extra 2 in it! (A)

I have a quick question about the small angle approximations. Why can you approximate that Cos~1 while you use the geometry for Sin rather than saying sin~theta?

JoshWasHere2014-08-20 08:20:22

When you expand cosine and sine, you see that the first two terms in the cosine and sine expansion are cos = 1 - and sin = - . Because of the small angle, we select only the first order terms, and drop all higher order terms (with the condition that < 5 degrees).

Energy conservation anyone? The change in gravitational potential energy should equate to potential between the charges. So... should equate to... hmmm well I usually prefer the energy method but the force way seems to be more effecient here.

Dimensional analysis helps!
Since is in N, will have units of
Mg has units of N
so will have units of
This leaves only A and B. Knowing that solving for will give half of d, take the solution with an extra 2 in it! (A)

It seems easier to equate torques around the pivot point, where the torque on the right mass due to the repulsive electric force isrnrn counterclockwisernrnand the torque on the right mass due to its weight is:rnrn clockwisernrn and rnrnFrom this, choice A follows.